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pishuonlain [190]

3 years ago

5

A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo

ut 61.0% of the roof. A typical annual rainfall is about 14.0 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water in a full tank of that size weigh (in N and in lbf)?

1 answer:

Kazeer [188]3 years ago

7 0

Answer:

volume = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

$volume = A'\times H$

where A' = 61% of A

$= 0.61\times 2667 = 1626.87 sq ft$

$volume = 1626.87 \times (\frac{14}{12} ft)$

=1898.015 ft^3

$in\ m^3 = \frac{ 1898.015}{35.315} = 53.7457 m^3$

$in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon$

$weight = \rho Vg$

$= 1000\times 53.74\times 9.8$

=526652 N

$In\ lbf = \frac{526652}{4.448} = 118396.08 lbf$

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2 years ago

A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i

Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

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Heat transfer per foot length is given by

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Put all the values in equation (1) , we get

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3 years ago

At an auction of antiques, a bidder for a particular porcelain statue would be trying to ________(a) Maximize the difference bet

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3 years ago

) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p

andrey2020 [161]

Answer:

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Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

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solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

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3 years ago

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

Answer:

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$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

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So;

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By solving:

$t_1 = 4.32 \ sec$

Thus;

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$\mathbf{Thus \ , the \ distance \ from \ the \ home \ plate \ = \ 299.36 \ feet}$

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